1997 USAMO Problems/Problem 2

Problem

$\triangle ABC$ is a triangle. Take points $D, E, F$ on the perpendicular bisectors of $BC, CA, AB$ respectively. Show that the lines through $A, B, C$ perpendicular to $EF, FD, DE$ respectively are concurrent.

Solution 1

Let the perpendicular from A meet FE at A'. Define B' and C' similiarly. By Carnot's Theorem, The three lines are concurrent if

$FA'^2-EA'^2+EC'^2-DC'^2+DB'^2-FB'^2 = AF^2-AE^2+CE^2-CD^2+BD^2-BF^2 = 0$

But this is clearly true, since D lies on the perpendicular bisector of BC, BD = DC.

QED

Solution 2

We split this into two cases: \\

Case 1: $D,E,F$ are non-collinear

Observe that since $D,E,F$ lie on perpendicular bisectors, then we get that $DC=DB$ , $EC=EA$ , and $FA=FB$ . This motivates us to construct a circle $\omega_1$ cantered at $D$ with radius $DC$ , and similarly construct $\omega_2$ and $\omega_3$ respectively for $E$ and $F$ .

Now, clearly $\omega_1$ and $\omega_2$ intersect at $C$ and some other point. Now, we know that $DE$ is the line containing the two centers. So, the line perpendicular to $DE$ and through $C$ must be the radical axis, which is exactly the line that the problem describes! We do this similarly for the others pairs of circles.

Now by the radical lemma, the pairwise radical axes of $\omega_1,\omega_2,\omega_3$ are concurrent, as desired, and they intersect at the radical center. \\

Case 2: $D,E,F$ are collinear

Now, we are drawing perpendicular lines from $A$ , $B$ , and $C$ onto the single line $DEF$ . Clearly, these lines are parallel and are never concurrent.

1997 USAMO (Problems • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5 • 6
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1997 USAMO Problems/Problem 2

Contents

Problem

Solution 1

Solution 2

See Also